# How to prove solutions of $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$?

The original question wanted the real solutions to $$2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 .$$ I graphed this equation and worked out the trivial roots $$x=-1$$ and $$x=1$$, and I could not find any more. Checking again, these are the only two solutions. So the problem is, I don't know how to write a formal derivation of those roots.

• Try factoring $(2^x-1)2$ on the RHS and dividing through on both sides. Also, notice that $2^{x^2}-2 = 2(2^{(x+1)(x-1)}-1)$. I don’t know if this works, but it’s worth a shot. – klein4 Apr 18 at 3:25
• The original question --- Giving us a clue about where this question appeared will help in determining what kinds of mathematical methods and what level of mathematical expertise is probably needed. For all we know, this could have been an unsolved research problem mentioned somewhere, or it's a math contest problem not requiring calculus (but perhaps a lot of ingenuity), or it's a transcendental equation that arose in some physics paper in which the author simply solved it by trial after inspecting a computer generated graph. – Dave L. Renfro Apr 18 at 3:30
• $x=0$ is another solution. – Josh Apr 18 at 3:30
• @CitrusCornflakes It is somewhat detailed, but I added a P.S. section containing the description of the natural generalisation of this type of an equation to ordered fields. Do take a look if you are interested. – ΑΘΩ Apr 18 at 12:45

Since for the right-hand term of the original equation we have $$2^{x+1}-2=2\left(2^x-1\right)$$, the equation can be rewritten as: $$2\left(2^x-1\right)\left(x^2-1\right)+2\left(2^{x^2-1}-1\right)x=0,$$ which after simplifying the factor $$2$$ reduces to the even simpler form: $$\left(2^x-1\right)\left(x^2-1\right)+\left(2^{x^2-1}-1\right)x=0.$$ It is easy to ascertain that the roots of the equations $$x=0$$ respectively $$x^2-1=0$$ (in other words, the elements of the set $$\{-1, 0, 1\}$$) are also roots of our equation. We proceed to show they are the only roots, by assuming the existence of an additional real root $$a \in \mathbb{R} \setminus \{-1, 0, 1\}$$ of our equation, other than these special values. In this case, via dividing by $$a\left(a^2-1\right)\neq0$$ the equation can be brought to the form: $$\frac{2^a-1}{a}+\frac{2^{a^2-1}-1}{a^2-1}=0.$$ We remark that for any $$y \in \mathbb{R}^{\times}=\mathbb{R} \setminus \{0\}$$ and and any $$t \in (1, \infty)$$ the inequality $$\frac{t^y-1}{y}>0$$ holds, so by the particular application of this observation for $$t=2$$ and $$y=a$$ respectively $$y=a^2-1$$ we gather that the left-hand side of the last equation above is strictly positive, which constitutes a contradiction. This concludes the argument.

P.S. There is a very natural way to immediately generalise this exercise and for this we require the following ingredients (objects):

• a commutative totally ordered field $$(K, +, \cdot, R)$$, in other words a commutative field $$(K, +, \cdot)$$ where the additional entity $$R$$ is a total order on the support set $$K$$ that is also compatible with the operations. Since the order is total, any element $$x \in K$$ satisfies only one of the relations $$x<_R 0_K$$, $$x=0_K$$, $$x>_R 0_K$$ and we go on to define the sign function: \begin{align*} \mathrm{sgn} \colon K &\to \left\{-1_K, 0_K, 1_K\right\} \\ \mathrm{sgn}(x)&=\begin{cases} -1_K, &x<_R 0_K \\ 0_K, &x=0_K \\ 1_K, &x>_R 0_K. \end{cases} \end{align*}
• for any arbitrary set $$M$$ we introduce the abbreviation $$\mathrm{T}(M)\colon=\mathrm{End}_{\mathbf{Ens}}(M)$$ for the collection of all self-maps of $$M$$ (which carries a natural monoid structure with respect to composition of maps, the so-called transformation monoid of set $$M$$). Given arbitrary self-map $$f \in \mathrm{T}(M)$$ we say subset $$X \subseteq M$$ is stable under $$f$$ if $$f[X] \subseteq X$$ and we write: $$\mathscr{S}\mathscr{tab}(f)\colon=\{X \subseteq M \mid f[X] \subseteq X\}$$ for the collection of all subsets stable under $$f$$. Returning to the commutative totally ordered field introduced above, let us write: $$\mathrm{S}_R(K)\colon=\{f \in \mathrm{T}(K) \mid (\leftarrow, 0_K)_R, \{0_K\}, (0_K, \rightarrow)_R \in \mathscr{Stab}(f)\}$$ for the collection of all sign-preserving self-maps of $$K$$ (where of course the sign is determined with respect to the fixed order relation $$R$$). More explicitly, a sign-preserving map will map $$0_K$$ to itself respectively strictly positives (negatives) to themselves. Note the equivalence: $$f \in \mathrm{S}_R(K) \Leftrightarrow f \in \mathrm{T}(K) \wedge \mathrm{sgn} \circ f=\mathrm{sgn}.$$
• arbitrary sets $$A$$ and $$I\neq \varnothing$$, where the latter is additionally required to be finite. Furthemore consider two families of maps, $$f \in \mathrm{Hom}_{\mathbf{Ens}}(A, K)^I$$ (this is simply a family of maps from $$A$$ to $$K$$ indexed by $$I$$) and $$g \in \mathrm{S}_R(K)^I$$ (this being a family of sign-preserving maps of $$K$$, again indexed by the same $$I$$).

If we define the map: $$h=\sum_{i \in I}\left(\left(g_i \circ f_i\right)\prod_{j \in I \setminus \{i\}}f_j\right),$$ then the roots of $$h$$ are described by the relation: $$h^{-1}[\{0_K\}]=\displaystyle\bigcup_{i \in I}f_i^{-1}[\{0_K\}].$$ In the particular case of your exercise we have $$K=\mathbb{R}$$ with the standard ordered field structure, $$A=\mathbb{R}$$, $$I=\{1, 2\}$$, $$g_1=g_2$$ given by $$g_1(x)=2^x-1$$ (in general the map $$x \mapsto a^x-1$$ will be sign-preserving as long as $$a>1$$) and finally $$f_1=\mathbf{1}_{\mathbb{R}}$$ respectively $$f_2(x)=x^2-1$$ for all $$x \in \mathbb{R}$$.

Partial solution: Here is an approach which didn't really pan out completly:

Rewrite as : $$2^{x}(x^2 - 1) + x2^{x^2 - 1} = x^2 + x - 1$$

Let $$y = x^2 - 1$$, then we get: $$\frac{y\cdot 2^x + x \cdot 2^y}{x+y} = 1$$ as we can check that $$x + y \neq 0 \Leftrightarrow x \neq \phi, -\frac{1}{\phi}$$.

For the case $$x > 1 \Rightarrow x > 0 \land y > 0$$, we have by AM-GM, there is no solution. I'm not sure how to deal with the case of $$x < 1$$ i.e. either of $$x < 0$$ or $$y < 0$$.